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Yuri [45]
2 years ago
8

A man pushes a shopping cart across a level floor. What force resists the effort force? A) gravity B) friction C) the normal for

ce D) the reaction force
Physics
1 answer:
Kipish [7]2 years ago
8 0

Answer:

B) Friction

Explanation:

Friction is a force that acts when an object is sliding along a surface. Microscopically, this force is due to the fact that the two surfaces are not perfectly smooth, but they have "imperfections" that cause a force that opposes the motion of the object.

For an object sliding on a flat surface, the force of friction has magnitude:

F_f = \mu_k mg

where

\mu_k is the coefficient of kinetic friction

m is the mass of the object

g is the acceleration of gravity

The direction of the force of friction is always opposite to the direction of motion of the object.

In reality, friction also acts if the object is at rest and it is pushed by a force; in this case, we talk about static friction, and its magnitude is

F_f = \mu_s mg

where \mu_s is called coefficient of static friction, and it is generally larger than the coefficient of kinetic friction.

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Please help me! As quickly as possible
k0ka [10]

Answer:

1. matter

2. kilograms

3. same

4. gravitational

5. gravity

6. space

7. weightlessness

8. Newton

9. weight

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I HOPE THESE ARE CORRECT AND IT HELPS

7 0
2 years ago
A race car accelerates from 0 m/s to 30.0 m/s with a displacement of
worty [1.4K]

Answer:

4. 10.0 m/s²

Explanation:

I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:

1. \ a=\frac{v-v_0}{t};

2. \ L=\frac{at^2}{2}.

II) using these two equations after substitution v₀=0; v=30 and L=45:

\left \{{{45 =\frac{at^2}{2}} \atop {a=\frac{30-0}{t} }} \right.

\left \{ {{at^2=90} \atop {at=30}} \right. \  \ \left \{ {{a=10} \atop {t=3}} \right. => \ a=10\frac{m}{s^2}

6 0
3 years ago
A 100 kg person rides in an elevator moving at a constant speed of
Airida [17]

Answer:

Explanation:

All this information only applies to the person. There is an extra tension force if we are talking about the elevator, but we are not. Dont forget to apply the units

Acceleration means change in speed or velocity. The elevator is moving at a constant speed of 3 meters. You wont even know you are moving because there is no change in acceleration. It equals 0

The forces ONLY acting on the person would be the force of gravity pulling them down, and the normal force that the elevator is reciprocating from the person standing on it.

Force = mass x acceleration. You have 100 kg and you are accelerating at 0 m/s. The force is 0. Which makes sense because the force of gravity and the net force completely cancel each other out.

8 0
3 years ago
What influences the path that surface currents take?
Anastasy [175]
The answer is wind forces and Earth’s rotation
8 0
3 years ago
Read 2 more answers
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
3 years ago
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